The procedures for computing real and complex solutions of one or more equations are shown in four tables on this page. Each procedure is broken down into steps, and each step includes an example calculation and a cell reference to sample spreadsheets. These sample sheets are shown on separate HTML files whose links appear in the top row of each procedure-table (except for Procedure 1, real solutions of one equation, whose sample sheet is shown below Procedure 1). If you have an advanced version of Excel, then you can copy rows of the sample sheets and paste them directly into an Excel file as active functions.

If your browser recognizes frames, then selecting the "Frames Version" of a procedure displays the procedure in the top frame, and the sample sheet in the bottom frame, so you can view simultaneously a step in the procedure and the corresponding Excel function. For all browsers, selecting the "No Frames Version" will display the procedure on this page without displaying the sample sheet. You can still view the procedure and sample sheet together by opening the sample sheet in a separate window, and then resizing the windows or tiling them horizontally. For Procedure 1, the procedure and sample sheet are so short that you can view them together without two windows.

Procedure 1. Finding Real Solutions of 1 Equation | Frames Version | No Frames Version |

Procedure 2. Finding Complex Solutions of 1 Equation | Frames Version | No Frames Version |

Procedure 3. Finding Real Solutions of Several Equations | Frames Version | No Frames Version |

Procedure 4. Finding Complex Solutions of Several Equations | Frames Version | No Frames Version |

Steps (6 total) | Example | View Cells in Sheet 1A, 1B below this table |

1. Write the equation in optimal form, F(w) = 0. | ^{9} - 23w^{2} + 5w - 19 = 0 | (not entered) |

2. Let variable z be an approximation of solution w. Enter a value for z in your spreadsheet. | First estimate: z = 1. | Sheet 1A, Cell A2 |

3. Let F(z) be a function based on the equation. Compute F in the cell next to z. | ^{9} - 23z^{2} + 5z - 19 | Sheet 1A, Cell B2 |

4. Let S(z) be the slope of F(z), | S(z) = 9z^{8} - 46z + 5
| Sheet 1A, Cell C2 |

5. Compute a new approximation with | Sheet 1A, Cell A3 | |

6. Repeat computations: copy all functions down their columns until z converges. | Solution: | Sheet 1B, Cell A9 |

Sheet 1A. Excel functions that compute real solutions of w^{9} - 23w^{2} + 5w - 19 = 0

A
B
C
1
Z
F
S
2
= 1
= A2^9 - 23*A2^2 + 5*A2 - 19
= 9*A2^8 - 46*A2 + 5
3
= A2 - B2/C2
= A3^9 - 23*A3^2 + 5*A3 - 19
= 9*A3^8 - 46*A3 + 5
4
= A3 - B3/C3
= A4^9 - 23*A4^2 + 5*A4 - 19
= 9*A4^8 - 46*A4 + 5

Sheet 1B. Values returned by functions in Sheet 1A.

A
B
C
1
Z
F
S
2
1
-36
-32
3
-0.125
-19.984375
10.7500005
4
1.734011536
62.2457253
660.863668
5
1.639823085
13.0895851
400.131897
6
1.607109909
1.14824921
331.577204
7
1.603646917
0.01161858
324.884311
8
1.603611155
1.2263E-06
324.815732
9
1.603611151
0
324.815725
10
1.603611151
0
324.815725

Steps (7 total) | Example | View Cells in |

1. Write the equation in optimal form, | ^{9} - 23w^{2} + 5w - 19 = 0 | (not entered) |

2. Let z = x + iy be an estimate of solution iv. | First estimate: x = 1, y = 1 | Sheet 2A A2:B2 |

3. Compute the absolute value R and imaginary argument q in the cells next to x and y | R = (x^{2} + y^{2})^{1/2}> 0, | Sheet 2A C2:D2 |

4. Let F(z) be a function based on the equation. Resolve F into real and imaginary parts, iFi. | F(z) = z^{9} - 23z^{2} + 5z - 19 = Fr + iFi
Fr = R ^{9}Cos(9q) - 23R^{2}Cos(2q) + 5RCos(q) -19
Fi = R ^{9}Sin(9q) - 23R^{2}Sin(2q) + 5RSin(q)
| Sheet 2A E2:F2 |

5. Let S(z) be the slope of F(z), iSi. | S(z) = 9z^{8} - 46z + 5 = Sr + iSi
Sr = 9R ^{8}Cos(8q) - 46RCos(q) +5
Si = 9R ^{8}Sin(8q) - 46RSin(q)
| Sheet 2A G2:H2 |

6. Compute a new approximation of iv^{2} + Si^{2})^{2} + Si^{2}) | Compute u and v in the cells below first estimate x and y. Use cell references in the approximation functions. | Sheet 2A A3:B3 |

7. Repeat computations: copy all functions down their columns until x and y converge. | Solution: w = 1.00541938 + i1.16519460
| Sheet 2B A9:B9 |

Steps (8 total) | Example (2 Equations, 2 Variables) | View Cells in |

1. Write the equations in optimal form,
_{1}(w_{1}, w_{2},...) = 0,_{2}(w_{1}, w_{2},...) = 0, ...
| w_{1}^{3}w_{2}^{5} - 6w_{1} w_{2} - 17 = 0
7w _{1}^{2}w_{2} - 3w_{1} w_{2}^{3} + 5 = 0
| (not entered) |

2. Let column-vector Z = [z_{1}, z_{2},...] be an approximation of solution W = [w_{1}, w_{2},...]. Assign values to Z for the first estimate. Enter Z as a column-vector in your spreadsheet.
| First estimate: Z = [1, 2]. | Sheet 3A A2:A3 |

3. Let column-vector F = [F_{1}(Z), F_{2}(Z),...] be a set of functions based on the equations. Compute F in the cells next to Z.
| F_{1}(Z) = z_{1}^{3}z_{2}^{5} - 6z_{1} z_{2} - 17
F _{2}(Z) = 7z_{1}^{2}z_{2} - 3z_{1} z_{2}^{3} + 5
| Sheet 3A B2:B3 |

4. Let J be the Jacobian matrix whose elements are the partial derivatives of F. Compute J in the cells next to F. | dF_{1}/dz_{1} = 3z_{1}^{2}z_{2}^{5} - 6z_{2}, dF_{1}/dz_{2} = 5z_{1}^{3}z_{2}^{4} - 6z_{1}
dF _{2}/dz_{1} = 14z_{1} z_{2} - 3z_{2}^{3}, dF_{2}/dz_{2} = 7z_{1}^{2} - 9z_{1} z_{2}^{2}
| Sheet 3A C2:D3 |

5. Let matrix 1/J be the inverse of J. Compute 1/J in the cells next to J. | In Excel, the element of 1/J in Row R, Column C is
= INDEX (MINVERSE (Matrix J), R, C) | Sheet 3A E2:F3 |

6. Let column-vector DZ be defined
| In Excel, the element of DZ in Row R is
= INDEX (MMULT (Matrix 1/J, Vector F), R, 1) | Sheet 3A G2:G3 |

7. Compute a new approximation using | Compute vector W in the cells below vector Z. | Sheet 3A A5:A6 |

8. Repeat computations: Copy the functions downward until Z converges and F approaches (0,0,...). | Solution: W = [1.110552, 1.845165] | Sheet 3B A14:A15 |

Steps (15 total) | Example (2 Equations, 2 Variables) | View Cells In |

1. Write the equations in optimal form,
_{1}(w_{1}, w_{2},...) = 0,_{2}(w_{1}, w_{2},...) = 0, | w_{1}^{3}w_{2}^{5} - 6w_{1} w_{2} - 17 = 0
7w _{1}^{2}w_{2} - 3w_{1} w_{2}^{3} + 5 = 0
| not entered |

2a. Let column-vector Z = [z_{1}, z_{2},...] be an approximation of solution _{1}, w_{2},...]. | First estimate:
i, 1 - i] | not entered |

2b. Resolve Z into real and imaginary parts,
iY. | Z = [1, 1] + i[-2, -1]
X = [1, 1] and Y = [-2, -1] | Sheet 4A A2:B3 |

3. Let column-vector R = [R_{1}, R_{2},...] be the absolute values of z_{1}, z_{2}, ... Compute R in the cells next to X and Y.
| For X = [x_{1}, x_{2}] and Y = [y_{1}, y_{2}],
R _{1} = (x_{1}^{2} + y_{1}^{2})^{1/2} and R_{2} = (x_{2}^{2} + y_{2}^{2})^{1/2}
| Sheet 4A C2:C3 |

4. Let column-vector q = [q_{1}, q_{2},...] be the imaginary arguments of z_{1}, z_{2}, ... Compute q in the cells next to R.
| q_{1} = + Arccos (x_{1}/R_{1}) if y_{1} > 0, and q _{1} = - Arccos (x_{1}/R_{1}) if y_{1} < 0.
q _{2} = + Arccos (x_{2}/R_{2}) if y_{2} > 0, and q _{2} = - Arccos (x_{2}/R_{2}) if _{2} < 0.
| Sheet 4A D2:D3 |

5a. Let column-vector _{1}(Z), F_{2}(Z),...] | F_{1}(Z) = z_{1}^{3}z_{2}^{5} - 6z_{1} z_{2} - 17
F _{2}(Z) = 7z_{1}^{2}z_{2} - 3z_{1} z_{2}^{3} + 5
| not entered |

5b. Resolve F into real and imaginary parts, iFi. | Elements of vector Fr:
Fr _{1} = R_{1}^{3}R_{2}^{5}Cos(3q_{1}+5q_{2}) - 6R_{1}R_{2}Cos(q_{1}+q_{2}) -17
Fr _{2} = 7R_{1}^{2}R_{2}Cos(2q_{1}+q_{2}) - 3R_{1}R_{2}^{3}Cos(q_{1}+3q_{2}) +5
Elements of vector Fi:
| Sheet 4A E2:E3
Sheet 4A |

6a. Let J be the Jacobian matrix whose elements are the partial derivatives of vector F. | dF_{1}/dz_{1} = 3z_{1}^{2}z_{2}^{5} - 6z_{2}, dF_{1}/dz_{2} = 5z_{1}^{3}z_{2}^{4} - 6z_{1}
dF _{2}/dz_{1} = 14z_{1} z_{2} - 3z_{2}^{3}, dF_{2}/dz_{2} = 7z_{1}^{2} - 9z_{1}^{2}z_{2}^{2}
| not entered |

6b. Resolve J into real and imaginary parts, iJi. | Elements of matrix Jr:
(dF _{1}/dz_{1})r = 3R_{1}^{2}R_{2}^{5}Cos(2q_{1} + 5q_{2}) - 6R_{2}Cos(q_{2})
(dF _{1}/dz_{2})r = 5R_{1}^{3}R_{2}^{4}Cos(3q_{1}+4q_{2}) - 6R_{1}Cos(q_{1})
(dF _{2}/dz_{1})r = 14R_{1} R_{2}Cos(q_{1}+q_{2}) - 3R_{2}^{3}Cos(3q_{2})
(dF _{2}/dz_{2})r = 7R_{1}^{2}Cos(2q_{1}) - 9R_{1}^{2} R_{2}^{2}Cos(2q_{1}+2q_{2})
Elements of matrix Ji:
| Sheet 4A
Sheet 4A |

7. Let matrices 1/Jr and 1/Ji be the inverses of Jr and Ji. Compute 1/Jr and 1/Ji in the cells next to Ji. | In Excel, the element of 1/Jr in Row R, Column C is
| Sheet 4A K2:L3 M2:N4 |

8. Let matrix Marie be defined
| In Excel, the element of Marie in Row R, Column C is
| Sheet 4A O2:P3 |

9. Let matrix 1/Marie be the inverse of Marie. Compute 1/Marie in the cells next to Marie. | In Excel, the element of 1/Marie in Row R, Column C is | Sheet 4A Q2:R3 |

10. Let column-vector Vince be defined
| In Excel, the element of Vince in Row R is
= INDEX(MMULT(Matrix 1/Ji, Vector Fr),R,1) + INDEX(MMULT(Matrix 1/Jr, Vector Fi),R,1) | Sheet 4A S2:S3 |

11. Let column-vector Vito be defined
| In Excel, the element of Vito in Row R is
= INDEX (MMULT (Matrix 1/Ji, Vector Fi), R, 1) - INDEX (MMULT (Matrix 1/Jr, Vector Fr), R, 1) | Sheet 4A T2:T3 |

12. Let column-vector DX be defined
DX = 1/Marie * Vince. Compute DX in the cells next to Vito. | In Excel, the element of DX in Row R is
| Sheet 4A U2:U3 |

13. Let column-vector DY be defined | In Excel, the element of DY in Row R is
= INDEX (MMULT (1/Marie, Vito), R, 1) | Sheet 4A V2:V3 |

14. Let W = U + iV. Compute the next approximation of W with
| Compute vectors U and V in the cells below X and Y. | Sheet 4A A5:A6 B5:B6 |

15. Repeat computations: Copy the rows of functions downward until X and Y converge, and Fr and Fi approach (0,0,...). | U = [0.396336, 1.201338]
V = [-1.259594, -0.891433] i1.259594, 1.201338 - i0.891433] | Sheet 4B A20:B21 |

URL: http://members.aceweb.com/patrussell/approximations/procedures.htm

Unpublished Work. © Copyright 2001 Pat Russell. Updated April 17, 2009.